Ch.3 - Chemical ReactionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Elemental phosphorus occurs as tetratomic molecules, P4. What mass (g) of chlorine gas is needed to react completely with 455 g of phosphorus to form phosphorus pentachloride?

Solution: Elemental phosphorus occurs as tetratomic molecules, P4. What mass (g) of chlorine gas is needed to react completely with 455 g of phosphorus to form phosphorus pentachloride?

Problem

Elemental phosphorus occurs as tetratomic molecules, P4. What mass (g) of chlorine gas is needed to react completely with 455 g of phosphorus to form phosphorus pentachloride?

Solution

We will begin solving this problem by analyzing what information we have, to solve this problem: 

Firstly, we will need to write a balanced equation for the data provided.

Phosphorous P4 reacts with Chlorine gas to form Phosphorous Pentachloride

Recall that, Chlorine gas - Cl2 exists as a diatomic molecule and the prefix -penta refers to 5 therefore, we can say that the following reaction takes place  

P4 (s) + Cl2 (g) → PCl5 (s)

  • Write down the number of atoms on the left side and the right side separately.
  • There are 4 P and 2 Cl on the left and 1 P and 5 Cl on the right.
  • Multiply PCl5  by 4 to have an equal number of P’s (4 and 4, therefore, they are equal).
  • Finally, there are 20 Cl on the right side and only 2 on the left. Multiply Clby 10. Now both sides have an equal number of atoms.
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